Answer:
146.3g NaCl (mol NaCl/58.44g NaCl) = 2.50 mol NaCl
1.5M NaCl = 1.5 mol NaCl / 1 L = 2.5 mol NaCl / x L, solve for x
x L = 2.5 mol NaCl / 1.5 mol NaCl = 1.66 L
It gives the answer and all the working.
To put it another way:
Dividing the amount required by the molar mass
we quickly see that 2.5 moles are required.
One litre of 1.5 molar solution gives 1.5 moles
we need a further mole, which is 2/3 of 1.5 so 2/3 of a litre.
Potential energy, kinetic energy would be if they were already running
Write a balance equation for the reaction between the analyte and the titrant.
Calculate the # of moles of titrant using the volume of titrant required and the concentration of titrant.
Calculate the # of moles of analyte using the stoichiometric coefficients of the equation.
Calculate the concentration of the analyte using the number or moles of analyte and the volume of analyte titrated.
Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= 
u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
The correct answer to your question is noble gases are stable <span>due to having the maximum number of valence electrons their outer shell can hold. Meaning their outer shells are stable.
Hope this helps let me know!</span>
